A reverse alphabet decreasing triangle prints each row backward down to A, while the starting letter moves one step earlier each row — so the left edge slides inward.
Remember
Rule: for start letter i from last down to A,
print i down through A
EDCBA
DCBA
CBA
BA
A ← 5 rows (every row ends at A)
Same widths as Program 6 (5, 4, 3, 2, 1), but Program 6 prints forward to a fixed end (ABCDE, BCDE, …). Here both loops decrement. Compare with Program 8, which keeps E fixed on the left and shortens the tail.
Approach
How to Solve It
Two ways to emit the same shape — start with nested ord/chr loops, then optionally reverse a prefix once and slice.
Method
Idea
Best for
Nested ord/chr loops
Outer = lowering start; inner = start..A downward
Learning, interviews, exams
Reverse + slice
Build EDCBA… once, drop one left letter each row
Shorter demos once loops click
Pseudocode
Pseudocode
for start from lastLetter down to 'A':
for ch from start down to 'A':
print ch (no newline)
print newline
Cheat sheet
Goal
Pattern
Lower start letter
for i in range(top, ord('A') - 1, -1):
Print i..A reverse
for j in range(i, ord('A') - 1, -1): print(chr(j), end="")
Print letters without a newline, then end the row once.
Try it
Live Preview
Change the row count and the reverse decreasing triangle updates instantly — capped at 26 letters (A–Z).
Whole numbers from 1 to 26. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 letters
EDCBA
DCBA
CBA
BA
A
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop start letter as i counts down from ord('D') to ord('A').
Start i
Inner j
Printed row
Letters
'D'
D..A
DCBA
4
'C'
C..A
CBA
3
'B'
B..A
BA
2
'A'
A..A
A
1
Total letter prints: 4 + 3 + 2 + 1 = 10 = 4×5/2 — same triangular count as Programs 1 and 6.
Code
Python Programs
Three complete programs: fixed top letter, input(), and a reverse-slice shortcut. Use View Output to reveal sample results.
Example 1 — Fixed from 'E' down to 'A'
Hard-coded top letter — both loops decrement; every row ends at A.
Python
for i in range(ord('E'), ord('A') - 1, -1):
for j in range(i, ord('A') - 1, -1):
print(chr(j), end="")
print()
Output
EDCBA
DCBA
CBA
BA
A
How It Works
1. Outer loop picks the start letter.i runs from ord('E') down to ord('A') — longest row first.
2. Inner loop walks down to A. For each start, j runs from i down to ord('A'), so the row is i..A in reverse.
3. Print letters, then break the line.print(chr(j), end="") stays on the row; bare print() after the inner loop starts the next (shorter) row.
When i is 'E' you get EDCBA; when i is 'A' you get A.
Example 2 — User Input Version
Read the row count at runtime. Prefer try/except ValueError and clamp to 26 (shown in the tip below).
Python
raw = input("Enter the number of rows: ").strip()
rows = int(raw)
top = ord('A') + rows - 1
for i in range(top, ord('A') - 1, -1):
for j in range(i, ord('A') - 1, -1):
print(chr(j), end="")
print()
Output (when user enters 4)
Enter the number of rows: 4
DCBA
CBA
BA
A
How It Works
1. Prompt and read. Ask for a row count, then convert the line to an int.
2. Map rows to a top letter.top = ord('A') + rows - 1 — for rows = 4, top is ord('D').
3. Same reverse-shrink core. Only the source of top changes — the print logic matches Example 1.
4. Safer input tip. Bare int(...) raises on letters. Prefer:
Safer input
try:
rows = int(input("Enter the number of rows: ").strip())
except ValueError:
print("Enter a whole number from 1 to 26.")
else:
if rows < 1 or rows > 26:
print("Enter a whole number from 1 to 26.")
else:
# draw pattern here
pass
Example 3 — Reverse + slice
Build the first reverse row once, then drop one left letter each time — same shape, no nested letter loop.
Python
rows = 5
letters = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
top = letters[:rows][::-1]
for i in range(rows):
print(top[i:])
Output
EDCBA
DCBA
CBA
BA
A
How It Works
1. Take and reverse the prefix.ABCDE reversed becomes EDCBA — the first printed row.
2. Slice from index i.top[0:] is the full row; top[1:] drops E; and so on down to A.
3. Learn loops first. Use Examples 1–2 when you need to show nested bounds; treat this as a polish shortcut afterward.
Edge Cases & Pitfalls
Check these before calling the solution done.
step +1
Program 6 by mistake
If the inner loop increments instead of decrements, you print forward suffixes. Use range(i, ord('A') - 1, -1).
Fixed start
Program 8 shape
If the outer loop stays on 'E' and only shortens the stop, you get EDCBA, EDCB, … Advance (lower) the start with a countdown outer range.
print early
Column of letters
If bare print() (or default print(chr(j))) is inside the inner loop, each letter lands on its own line. Use end="" for letters; call print() only after the inner loop.
rows > 26
Past Z
ord('A') + rows - 1 leaves A–Z when rows > 26. Clamp or reject in interactive programs.
rows = 1
Single A
Output is just A — start and end coincide. A good sanity check.
Bad input
Catch ValueError
Bare int(input()) raises on letters — wrap in try/except ValueError and require 1–26.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(rows²)
O(1)
Reverse + slice (Example 3)
O(rows²)
O(rows) for the reversed prefix and temporary row strings
Total letters = n + (n - 1) + … + 1 = n(n + 1)/2 — quadratic in n. Same totals as Programs 1 and 6.
Remember
Key Takeaways
Rule: lower start letter i; print i down through A each row.
vs Program 6: same widths — here letters run backward to A; there they run forward to a fixed end.
Break the row: call bare print() only after the inner loop.
Complexity:O(n²) time from the triangular letter count; O(1) extra space for nested loops.
One line: for i from the top letter down to 'A', print i down through A, then print().
Frequently Asked Questions
The outer loop picks the starting letter (E, then D, then C…). The inner loop prints from that start down to A. Each row drops the previous leftmost letter and becomes shorter.
The outer loop decreases the starting letter each row, and the inner loop decreases letters within that row down to A. That matches the reverse-alphabet output.
Because the inner range includes A: range(start, ord('A') - 1, -1). The stop is one below A so A is printed.
Program 6 prints forward suffixes ending at E (ABCDE, BCDE, …). This pattern prints reverse runs ending at A (EDCBA, DCBA, …). Same shrinking widths; opposite letter direction.
Program 8 always starts each row at E and shortens the tail (EDCBA, EDCB, EDC). Program 7 changes the first letter each row (E, then D, then C…).
print(..., end="") stays on the same line. print() ends the current line. Letters use end=""; the row break uses print() after the inner loop.
O(n²) for n rows, because total printed characters are n(n+1)/2.
Use try/except ValueError around int(input()) and clamp rows between 1 and 26 so bad input does not raise or walk past Z.
🤔
Did you know?
Both loops run backward: the outer loop lowers the start letter; the inner loop prints down to A. For 5 rows: EDCBA, DCBA, CBA, BA, A. Same widths as Program 6, opposite letter direction.
